Number Sequence

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

Sample Input

2
5
2 1 2 3 1 2 3 1 3 2 1 2
2 3 1 3
5
2 1 2 3 1 2 3 1 3 2 1 2
2 3 2 1

Sample Output

6
-1

Source

HDU 2007-Spring Programming Contest

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#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include <stdio.h>
#include <string.h>
using namespace std;
int a[1000009] ;
int b[100009];

void getnext(int *b , int len , int *next)//next是记录字符串的每个字符的之前的字符串的最长的前缀与后缀
{
    next[0] = -1;//将next数组右移一项使与查找时下标匹配
    int  j = 0  , k = -1 ;
    while(j < len - 1)
    {
        if(k == -1 || b[k] == b[j])
        {
            k++;
            j++;
            next[j] = k ;
        }
        else
        {
            k = next[k];
        }
    }
}

int main()
{
    int t ;
    cin >> t ;
    while(t--)
    {
        int next[100009];
        int n , m ;
        scanf("%d%d" , &n ,&m);
        for(int i = 0 ; i < n ; i++)
        {
            scanf("%d" , &a[i]);
        }
        for(int i = 0 ; i < m ; i++)
        {
            scanf("%d" , &b[i]);
        }
        getnext(b , m , next);//求next数组
        int i = 0 , j = 0;
        while(i < n && j < m)
        {
            if(j == -1 || a[i] == b[j])
            {
                j++ ;
                i++ ;
            }
            else
            {
                j = next[j];//文本i不动，b串移动到最长前缀与后缀的长度下标
            }
        }
        if(j == m)
            printf("%d
" , i - j + 1);
        else
            printf("-1
");


    }



    return 0 ;
}